1. Solve an equation with two real solutions
EXAMPLE 1
Solve an equation with two real solutions
Solve x2 + 3x = 2.
x2 + 3x = 2
Write original equation.
x2 + 3x – 2 = 0
Write in standard form.
x =
– b + b2 – 4ac 2a
Quadratic formula
x =
– – 4(1)(–2)
2(1)
a = 1, b = 3, c = –2
x =
– 3 + 17 2
Simplify.
The solutions are
x =
–3 + 17 2
0.56 and
–3 –
–3.56.
ANSWER

2. EXAMPLE 1
Solve an equation with two real solutions
CHECK
Graph y = x2 + 3x – 2 and note that the x-intercepts are about 0.56 and about –3.56. 

3. Solve an equation with one real solutions
EXAMPLE 2
Solve an equation with one real solutions
Solve 25x2 – 18x = 12x – 9.
25x2 – 18x = 12x – 9.
Write original equation.
25x2 – 30x + 9 = 0.
Write in standard form.
x =
(–30)2– 4(25)(9)
2(25)
a = 25, b = –30, c = 9
x = 50
Simplify.
x = 3 5
Simplify. 3 5
The solution is
ANSWER

4. EXAMPLE 2
Solve an equation with one real solutions
CHECK
Graph y = –5x2 – 30x + 9 and note that the only x-intercept is 0.6 = . 3  5

5. Solve an equation with imaginary solutions
EXAMPLE 3
Solve an equation with imaginary solutions
Solve –x2 + 4x = 5.
–x2 + 4x = 5
Write original equation.
–x2 + 4x – 5 = 0.
Write in standard form.
x =
– 42 – 4(–1)(–5)
2(–1)
a = –1, b = 4, c = –5
x =
– –4 –2
Simplify.
–4 + 2i
x = –2
Rewrite using the imaginary unit i.
x = 2 + i
Simplify.
The solution is 2 + i and 2 – i.
ANSWER

6. EXAMPLE 3
Solve an equation with imaginary solutions
CHECK
Graph y = 2x2 + 4x – 5. There are no x-intercepts. So, the original equation has no real solutions. The algebraic check for the imaginary solution 2 + i is shown.
–(2 + i)2 + 4(2 + i) = 5 ?
–3 – 4i i = 5 ?
5 = 5 

7. GUIDED PRACTICE
for Examples 1, 2, and 3
Use the quadratic formula to solve the equation.
x2 = 6x – 4
ANSWER
4x2 – 10x = 2x – 9 1 2
ANSWER 1
7x – 5x2 – 4 = 2x + 3
115
5 + i 10
ANSWER